Quadratic Diophantine Equations With xy Terms

We can solve Diophantine equations with mixed  
  terms by completing the square and transforming variables.
Example: Solve  
Completing the square gives  
. Let  
Convergents of the continued fraction representing  
\[\sqrt{3}= [ 1, \lt 1,2 \gt ]\]

\[1+ \frac{1}{1}=2\]

\[1+\frac{1}{1+ \frac{1}{2}}=\frac{5}{3}\]

\[1+\frac{1}{1+ \frac{1}{2+\frac{1}{1}}}=\frac{7}{4}\]

\[1+\frac{1}{1+ \frac{1}{2+\frac{1}{1+\frac{1}{2}}}}=\frac{19}{11}\]

\[1+\frac{1}{1+ \frac{1}{2+\frac{1}{1+\frac{1}{2+\frac{1}{1}}}}}=\frac{26}{15}\]

Solutions are given by then convergent for the penultimate element of the cycle part of the continued fraction - 1. These are the convergents  
\[\frac{2}{1}, \; \frac{7}{4}, \; \frac{26}{15}\]
\[\frac{2}{1} z=x+y=2, \; y=1 \rightarrow x=1, \; y=1\]

\[\frac{7}{4} z=x+y=7, \; y=4 \rightarrow x=3, \; y=4\]

\[\frac{26}{15} z=x+y=26, \; y=15 \rightarrow x=11, \; y=15\]

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